Suppose I have a matrix $$M$$ and $$||M||_2$$ denotes the spectral radius of the matrix.

I came across a note which says $$||M||_2 \leq \sqrt{||M||_1||M||_\infty}$$.

Could someone explain to me how this inequality holds? I don't understand it.

Let $$v$$ be a vector of length $$1$$.
$$||Mv||^2 = \sum_{i=1}^n \left(\sum_{j=1}^n m_{ij}v_j \right)^2 \\ \leq \sum_{i=1}^n \left(\sum_{j=1}^n m_{ij}^2 \right) \left(\sum_{j=1}^n v_j^2 \right) \\ = \sum_{i=1}^n \sum_{j=1}^n m_{ij}^2 = \sum_{i=1}^n \sum_{j=1}^n |m_{ij}| |m_{ij}| \\ \leq \sum_{i=1}^n \sum_{j=1}^n |m_{ij}| ||M||_{\infty} \\ = ||M||_1 ||M||_{\infty}$$
• In the last inequality we use the fact $$|m_{ij}| \leq ||M||_{\infty}$$ for all $$i,j$$.
• The last equality is the definition of $$||M||_1$$, since the constant $$||M||_{\infty}$$ can be pulled out of the sum.
$$||M||_2^2=\rho(M^*M)\leq||M^*M||_1\leq||M^*||_1||M||_1=||M||_{\infty}||M||_1$$